The Frequency of Summands of a Particular Size in Palindromic Compositions

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1 The Frequency of Summands of a Particular Size in Palindromic Compositions Phyllis Chinn Dept. of Mathematics, Humboldt State University, Arcata, CA phyllis@math.humboldt.edu Ralph Grimaldi Dept. of Mathematics, Rose-Hulman Institute of Technology Terre Haute, IN ralph.grimaldi@rose-hulman.edu Silvia Heubach Dept. of Mathematics, California State University Los Angeles 5151 State University Drive, Los Angeles, CA sheubac@calstatela.edu Abstract A composition of a positive integer n consists of an ordered sequence of positive integers whose sum is n. A palindromic composition is one for which the sequence is the same from left to right as from right to left. This paper shows various ways of generating all palindromic compositions, counts the number of times each integer appears as a summand among all the palindromic compositions of n, and describes several patterns among the numbers generated in the process of enumeration. Keywords: Compositions, palindromes, tilings. A.M.S. Classification Number: 05A99 1

2 1 Introduction A composition of a positive integer n consists of an ordered sequence of positive integers whose sum is n. It is well-known that there are 2 n 1 compositions of n (see for example [3]). A palindromic composition is one for which the sequence is the same from left to right as from right to left. For the remainder of this paper we will refer to such compositions by the short-hand term palindrome. Compositions can also be thought of as tilings ofa1xn board, with 1 x k tiles of integer length k, 1 k n. In this setting, a composition of n with j summands or parts is created by making j 1 vertical cuts on the 1 x n board. This viewpoint allows for easy combinatorial proofs of certain facts and will be used when advantageous. The question concerning the number of times a particular summand k occurs in all compositions of n has been answered by one of the authors in [3]. Furthermore, Chinn et al. showed that the number of times k appears as a summand in compositions of n is equal to the number of times k + 1 appears in compositions of n + 1. Alladi and Hoggatt enumerated the number of times the summands 1 and 2 occur in all compositions and palindromes containing only these two summands [1]. Grimaldi has investigated compositions with odd summands, and expressed the number of times a 1 occurs in all compositions of n with odd summands as a specific linear combination of Lucas and Fibonacci numbers [4]. Furthermore, the occurrence of the number 2k + 1 in all compositions of n with odd summands equals the number of 1s in all compositions of n 2k with odd summands. We will show a somewhat similar result for palindromes, namely that the number of times the summand k occurs in a palindrome of a specific size can sometimes be reduced to the number of 1s in all palindromes of a certain smaller size. In addition, the sequence of values of occurences of 1s in palindromes of even and odd values of n, respectively, matches known sequences (A and A in [7]). Section 2 contains notation and a few basic observations that will be used throughout the rest of the paper. In Section 3, we describe two methods of generating palindromes, and give a formula for the total number of palindromes. Section 4 contains explicit formulas for R n (k), the number of times the number k occurs as a summand among all the palindromes of n. We conclude in Sections 5 and 6 by discussing the various patterns found within the table of values for R n (k), and give combinatorial or analytical proofs for these patterns. 2

3 2 Notation and General Observations Before deriving specific results, we will define our notation, and state a remark which will be used in later sections. Let C n = the number of compositions of n, wherec 0 := 1 P n = the number of palindromes of n, wherep 0 := 1 R n (k) = the number of repetitions of the integer k in all palindromes of n. Remark 1 1. A palindrome of an odd integer n always has an odd number of summands, and the middle summand must be an odd integer. 2. A palindrome of an even integer n can have an odd number of summands with an even summand in the center or an even number of summands and no middle summand. We will refer to a palindrome of the latter type as having an even split. 3 Generating Palindromes Palindromes can be created in a number of ways, each of which is useful for some of the proofs in this section. In addition, these different creation methods illustrate the multiple ways of thinking about palindromes. The first method creates palindromes using compositions, whereas the second method creates palindromes recursively. We start by describing the explicit method of palindrome creation, which consists of combining all possible middle summands with a composition of an appropriate positive integer to the left, and with its mirror image on the right. This method will be referred to as the Explicit Palindrome Creation Method (EPCM): To create a palindrome of n =2k (n =2k +1), combine the middle summand m =2l (m =2l +1), forl =0,..., k, witha composition of n m 2 = k l on the left and its mirror image on the right. For those palindromes that result from l = 0, delete the middle summand of 0. The second method creates palindromes recursively; to seed this method, we define a palindrome of n = 0, namely 0. We will refer to this method as 3

4 the Recursive Palindrome Creation Method (RPCM): Before applying the algorithm, create a middle summand for palindromes with an even number of summands by replacing the + sign in the center of the palindrome by +0+. (This artifice simplifies the algorithm and allows the treatment of palindromes having an odd and even number of summands, respectively, using the same instructions.) 1. Creating palindromes of 2k +1 from those of 2k: Increase the middle summand by Creating palindromes of 2k +2 from those of 2k: Create one palindrome by increasing the middle summand by 2, and another one by replacing the middle summand m by ( m 2 +1) + ( m 2 +1). Lemma 2 for n 1. Both the EPCM and the RPCM create all palindromes of n Proof: Clearly, the EPCM creates all palindromes of n, without duplicates or omissions. For the RPCM, we need to work a little harder to show that indeed no duplicates are created, and also that all possible palindromes are created by the algorithm. For easier readability we will refer to the middle summand(s) of a palindrome of n as m n. Furthermore, we will only concentrate on the middle summands, as all other summands remain unchanged when creating the palindromes of 2k +1 and 2k +2, respectively, from those of 2k. Palindromes of 2k +1: Every palindrome of 2k + 1 with middle summand m 2k +1 corresponds to a palindrome of 2k whose middle summand is m 2k (If m 2k +1 = 1, then the corresponding palindrome of 2k is the one where the dummy 0 summand is deleted.) Palindromes of 2k + 2: No duplicates are created as distinct palindromes of 2k lead to distinct palindromes of 2k + 2 for each instruction. Furthermore, the first instruction creates palindromes with an odd number of summands, whereas the second instruction creates palindromes with an even number of summands. Thus, if a palindrome of 2k + 2 has an odd number of summands, then it is created from the palindrome of 2k whose middle summand is m 2k If, on the other hand, the palindrome of 2k + 2 has an even number of summands, then it is created from the palindrome of 2k whose middle summand is 2 (m 2k +2 1). (If m 2k =0,then delete the dummy 0 summand.) 4

5 Initial conditions: This algorithm creates the one palindrome of n = 1, namely 1, and the two palindromes of n = 2, namely 2 and 1 + 1, from the initial condition. The recursive method immediately shows some of the structure within the palindromes. Remark 3 1. The first rule of the RPCM demonstrates that half of the palindromes of an odd integer n have a 1 as the middle summand (since half of the palindromes of n 1 had a dummy zero summand). 2. The second rule of the RPCM illustrates that half of all the palindromes of an even integer n have an even number of summands. Using either the RPCM or the EPCM, we can easily determine the total number of palindromes of n. Theorem 4 For k 0, P 2k = P 2k+1 =2 k,wherep 0 := 1. Proof: In the RPCM, the number of palindromes stays the same when creating the palindromes of 2k + 1 from those of 2k, and the number of palindromes doubles when creating the palindromes of 2k + 2. Thus, P 2k+1 = P 2k and P 2k =2P 2(k 1) =2 2 P 2(k 2) =... =2 k 1 P 2 =2 k which completes the proof. 4 The Frequency of k in Palindromes of n The question regarding how many times the summand k appears among all the palindromes of n is motivated by the comparable question regarding compositions as explored in [3]. The following theorem is proved in that paper. Theorem 5 The number of repetitions of the integer k in all of the compositions of n is (n k +3) 2 n k 2 for n>kand 1 for n = k. 5

6 The following theorem states the corresponding result for palindromes. We need to consider different cases according to whether or not n and k have the same parity, and also according to the relative size of n and k. In particular, we get a different pattern when n is too small to accommodate two summands of k within a single palindrome. Theorem 6 For n<k, R n (k) =0.Ifnand k have different parity, then { 0 k<n<2k R n (k) = 2 n/2 k (2 + n 2 k) n 2k. If n and k have the same parity, then 1 n = k R n (k) = 2 (n k)/2 1 k<n<2k 2 n/2 k (2 + n k+1 2 k ) n 2k. Proof: Let n =2i or n =2i +1,andk =2j or 2j + 1, respectively. For n<k, the palindrome cannot contain the summand k. Ifn = k, then there is exactly one palindrome that contains the summand k, namelyjustk by itself. If k<n<2k, then the summand k can occur at most once in any palindrome, and hence has to occur in the center. This is only possible if n and k have the same parity (by Remark 1), which implies that R n (k) =0 if n and k have different parity. If they have the same parity, then the palindromes that have the summand k in the center can be created using the explicit method. Thus, the number of repetitions of k is given by the number of compositions of size ( ) n k 2 = i j, whichgivesrn (k) =2 i j 1. If n 2k, then the summand k can occur in the center, or in symmetric pairs at other positions within the palindrome. To count the different cases, we will think of the palindrome as a 1 x n board as illustrated in Figure n-1 n Figure 1: Palindrome as a 1 x n board We will count according to whether a tile of length k starts at position s, for 1 s i k + 1, as we will only look at the left half of the tiling. Tilings that contain a tile of size k starting at position s can be created by combining the tile of size k with any tiling (i.e., composition) 6

7 of length s 1 on the left, and a symmetric tiling (i.e., palindrome) of length n 2(s 1) 2k on the right, and then completing the remainder of the tiling symmetrically. If n and k have the same parity, we also get occurrences of k in the center. We look first at the case where n and k have different parity: i k+1 i k+1 R n (k) = 2 C s 1 P n 2(s+k 1) =2 C s 1 P 2(i s k+1) s=1 s=1 i k+1 = 2 C 0 P 2(i k) +2 2 s 2 2 i s k+1 s=2 = i k +2 2 i k 1 (i k) =2 i k (2 + i k) (1) which gives the formula for R n (k) forn 2k where n and k have different parity. Lastly, we consider the case where n and k havethesameparityand n 2k. In this case, the number of occurrences of k is given by off-center ones (as counted in Eq. (1)), plus those that occur in the center. The latter is given by C i j =2 i j 1 (see the case k<n<2k). Altogether, R n (k) = 2 i k (2 + i k)+2 i j 1 { 2 = i k (2 + i k +2 j 1 ) if k =2j 2 i k (2 + i k +2 j ) if k =2j +1 which proves the formula for the case n 2k where n and k have the same parity. These two cases can be written using a single formula by noting that k gives the correct powers of j 1andj, respectively. Table 1 displays the values of R n (k) that arise from the formulas given in Theorem 6. Examining the values in Table 1 led the authors to observe a variety of patterns. Some of these follow from combinatorial arguments while others just seem to be consequences of the formulas given in Theorem 6. In Section 5 we will present those patterns that hold across the table, and give combinatorial proofs for them. Patterns that hold only for specific columns will be discussed in Section 6. As before, we let n =2i or n =2i +1,andk =2j or k =2j + 1, respectively. 5 General Patterns in the Repetitions of k in all Palindromes of n The most striking pattern in the table is the equality of certain diagonally adjacent entries. Furthermore, diagonal sequences that start in column 1 7

8 n\k Table 1: The number of occurrences of k among all palindromes of n for n =2i are repeated on the diagonal that starts in row 2i +2,withtwo new entries inserted at the beginning of the lower diagonal. Note also that the values that occur on these diagonals are comprised of the values for even rows in column 1 (above the starting row for the diagonal), in reverse order. Theorem 7 a) R 2i+1 (2j) =R 2i+2 (2j +1) for i j 1. b) R 2i (2j 1) = R 2i+3 (2j), for i j 1. c) R 2i+2l (2l +1)=R 2i 2l (1) for l 1. Proof: a) To show the first equality, note that a palindrome of an odd integer n must have an odd middle summand; thus, no copy of 2j occurs in the center. For i j, pairs of (2j)s can occur. For each pair of symmetrically located occurrences of 2j in a palindrome of 2i +1, there is a 8

9 corresponding palindrome of 2i + 2 which has a pair of symmetrically located occurrences of 2j + 1 and whose middle summand is decreased by one. Since a palindrome of an even integer n cannot have 2j +1 as the middle summand, the number of occurrences of 2j in the palindromes of 2i + 1 equals the number of occurrences of 2j + 1 in the palindromes of 2i +2. b) To show the second equality, which together with part a) leads to the repeated diagonals, we make a similar argument. Since a palindrome of an even integer n must have an even middle summand (possibly 0), no copy of 2j 1 occurs in the center. For i j, pairs of (2j 1)s can occur. For each pair of symmetrically located occurrences of 2j 1in a palindrome of 2i, there is a corresponding palindrome of 2i +3 which has a pair of symmetrically located occurrences of 2j and whose middle summand is increased by 1. Since the palindrome of 2i + 3 cannot have an even summand in the center, there is a one-to-one correspondence between the occurrences of the (2j 1)s in the palindromes of 2i and the (2j)s in the palindromes of 2i +3. c) Both 2i +2l and 2i 2l are even, and we are counting the number of occurrences of 2l + 1 and 1, respectively. Neither of these can occur in the center of the palindromes. To make the association between the palindromes of the two sizes, we think of the palindrome as a symmetric tiling. For a tiling of length 2i 2l which has at least one pair of 1 x 1 tiles, replace one pair of 1 x 1 tiles with a pair of 1 x 2l tiles. This increases the length of the tiling to 2i 2l +2(2l) =2i +2l, and each pair of 1s in the shorter tiling has an associated pair of (2l + 1)s in the longer tiling. Thus, the number of 1s in the palindromes of 2i 2l equals the number of (2l + 1)s in the palindromes of 2i +2l. Figure 2 illustrates this process for i =3andl = 1 to show that R 8 (3) = R 4 (1). There are two palindromes of 4 that contain 1s: and 1+2+1, and 3 palindromes of 8 that contain 3s: , , and Figure 2: Replacing pairs of 1s by pairs of (2l +1)s 9

10 Note that a palindrome of 2i 2l with j pairs of 1s will have j palindromes of 2i + 2l associated with it. However, the correspondence of the pairs is one-to-one. For diagonals that start in column 1 in a row for odd n, weonly get equality of adjacent pairs, but not a repetition of the whole diagonal sequence. Theorem 8 R 2i+1 (2j 1) = R 2i+2 (2j) for i j 1. Proof: A palindrome of an odd integer must have an odd middle summand. If this middle summand is 2j 1, increase it by 1 to get a palindrome of 2i + 2 with middle summand 2j. Fori j, we also get symmetric pairs of (2j 1)s. Increase each 2j 1by1to2j, and decrease the middle summand by 1. Thus, there is a one-to-one correspondence between the occurrences of 2j 1 in the palindromes of 2i + 1 and the occurrences of 2j in the palindromes of 2i +2. The next pattern is a bit more complex. Theorem 9 The sum of two adjacent entries for even n in an appropriate set of two columns is equal to the sum of the two adjacent entries below them: R 2i (2j)+R 2i (2j +1)=R 2i+1 (2j)+R 2i+1 (2j +1) for i j 1. Proof: Consider any even palindrome. Using the RPCM, the palindromes of the next odd integer are generated by increasing the middle summand by 1. Note, however, that in half of the palindromes of 2i this middle summand is a dummy 0 and the increase therefore does not change the number of occurrences of any integer greater than 1; in particular the number of occurrences of 2j and 2j + 1 remains unchanged. In the other half of the palindromes of 2i, the middle summand is even and at least 2. Increasing a middle summand of size 2j leads to a loss in the count of (2j)s, which is, however, compensated for by an increase in the number of (2j +1)s. Before stating patterns that are specific to particular columns of Table 1, we will focus on the values of R n (1) for even and odd values of n, respectively. For k = 1, the formulas given in Theorem 6 simplify to R 2i (1) = (i +1) 2 i 1 and R 2i 1 (1) = (i +1) 2 i+1 for i 1. For even n, the sequence of values R 2i (1), given by {2, 6, 16, 40, 96, 224, 10

11 512, 1152, 2560, 5632, 12288,...}, matches the sequence a(i) defined in A of [7] (with R 2i (1) = a(i 1)),whicharisesasthenumberof states in a ferry problem [5]. For odd n, the sequence of values R 2i 1 (1), given by {1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144,...}, matches the sequence a(i) defined in A of [7] (with R 2i+1 (1) = a(i)). This sequence arises in several different contexts, for example in generalizations of the Stirling number triangles [6] and as a realization of oligomorphic permutation groups [2]. Now imagine that we color all the values that belong to a known sequence. Due to the repeated diagonals, the sequence for R 2i (1) occurs in all columns. If k is odd, the sequence occurs in the even rows, and if k is even, it occurs in the odd rows. The first non-zero value, 2, occurs for n =2k +1 when k is even, and for n =2k when k is odd. If the preceeding zeros are included, then these values fill all the diagonals that start in an even row in column 1, giving a checker-board coloring of the table. We consider the remaining uncolored sequences in each column. Intheevenrowsofcolumn2,wegetthesequenceforoddrowsofcolumn 1, due to the equality of diagonally adjacent entries, thus column 2 is now completely colored. Likewise, the remaining uncolored sequences in adjacent odd and even columns are the same. We tested these uncolored sequences, {4, 10, 24, 56, 128, 288, 640, 1408, 3072,...} (for columns 3 and 4), {6, 14, 32, 72, 160, 352, 768, 1664, 3584,...} (for columns 5 and 6), {10, 22, 48, 104, 224, 480, 1024, 2176, 4608,...} (for columns 7 and 8), {18, 38, 80, 168, 352, 736, 1536, 3200, 6656,...} (for columns 9 and 10), and {34, 70, 144, 296, 608, 1248, 2560, 5248, 10752,...} (for columns 11 and 12), both with and without the entries for n<2k, which are described by a different formula than those for n 2k, against the On-Line Encyclopedia of Integer Sequences [7]. (The sequences above list only the values for n 2k). The fact that none of these sequences occurs makes it unlikely that sequences for values of k 13 are in the encyclopedia; we are therefore in the process of submitting this family of related sequences to the encyclopedia. 6 Specific Patterns in the Repetitions of k in all Palindromes of n The remaining patterns are specific to particular columns of Table 1. We present only analytical proofs for these, rather than combinatorial ones. The fact that the patterns hold only for specific columns seems to indicate that no general method similar to those used in the proofs in Section 5 is applicable. For each of the following theorems, the range indicated for i ensures that for all values of n and k, n 2k holds. 11

12 Theorem 10 a) R 2i (1) = 2 R 2i+1 (2) + 2 i 1 for i 2. b) R 2i (1) = R 2i+2 (3) + R 2i+3 (3) for i 2. Proof: Using the appropriate formula in Theorem 6, we get: R 2i (1) = 2 i 1 (2 + i 1) = 2 i 1 (i +1), 2 R 2i+1 (2) + 2 i 1 = 2 (2 i 2 (2 + i 2)) + 2 i 1 =2 i 1 (i +1), and R 2i+2 (3) + R 2i+3 (3) = 2 (i+1) 3 (2 + (i +1) 3) +2 (i+1) 3 (2 + (i +1) ) = 2 i 2 (i + i +2)=2 i 1 (i +1), which completes the proof. Theorem 11 R 2i+1 (1) = R 2i+4 (3) + R 2i+3 (3) R 2i+2 (3) for i 1. Proof: From Theorem 6 we get: R 2i+1 (1) = 2 i 1 (2 + i )=2 i 1 (i +2) and R 2i+4 (3) + R 2i+3 (3) R 2i+2 (3) = 2 (i+2) 3 (2 + (i +2) 3) + 2 (i+1) 3 (2 + (i +1) ) 2 (i+1) 3 (2 + (i +1) 3) = 2 i 2 (2(i +1)+(2+i) i) =2 i 2 (2(i +2)+i i) = 2 i 1 (i +2), which proves the statement. Theorem 12 R 2i (2) = 2 R 2i+1 (3) for i 3. Proof: Again, we use the formula for R n (k) given in Theorem 6. R 2i (2) = 2 i 2 (2 + i )=2 i 2 (i +1) = 2 [ 2 i 3 (2 + i ) ] =2 R 2i+1 (3) which completes the proof. The next three theorems seem to have a similar structure, but there is no general underlying pattern. Furthermore, these types of pattern do 12

13 not seem to occur for larger values of k. The second pattern in Theorem 15 also differs somewhat from the ones of Theorems 13 and 14 in that the values are expressed as a difference rather than as a sum. Theorem 13 a) R 2i+1 (2) = 4 R 2i 1 (3) for i 4. b) R 2i+1 (2) = R 2i+2 (4) + R 2i+3 (4) for i 3. Proof: Using Theorem 6, and R 2i+1 (2) = 2 i 2 (2 + i 2) = 2 i 2 i, [ ] 4 R 2i 1 (3) = 4 2 (i 1) 3 (2 + (i 1) ) =2 i 2 i, R 2i+2 (4) + R 2i+3 (4) = 2 (i+1) 4 (2 + (i +1) ) +2 (i+1) 4 (2 + (i +1) 4) = 2 i 3 [(i +1)+(i 1)] = 2 i 2 i, which proves the desired equalities. Theorem 14 a) R 2i (3) = 4 R 2i 2 (4) for i 5. b) R 2i (3) = R 2i (4) + R 2i+1 (4) for i 4. Proof: The formulas for R n (k) intheorem6give R 2i (3) = 2 i 3 (2 + i 3) = 2 i 3 (i 1), [ ] 4 R 2i 2 (4) = 4 2 (i 1) 4 (2 + (i 1) ) =2 i 3 (i 1), and R 2i (4) + R 2i+1 (4) = 2 i 4 (2 + i )+2 i 4 (2 + i 4) = 2 i 4 (i + i 2) = 2 i 3 (i 1). This completes the proof. Theorem 15 a) R 2i (4) = 4 R 2i 1 (5) for i 6. b) R 2i (4) = R 2i+3 (4) R 2i+2 (5) for i 4. 13

14 Proof: Once more we use the formula for R n (k) given in Theorem 6. and R 2i (4) = 2 i 4 (2 + i )=2 i 4 i [ ] = 4 2 (i 1) 5 (2 + (i 1) ) =4 R 2i 1 (5) R 2i+3 (4) R 2i+2 (5) = 2 (i+1) 4 (2 + (i +1) 4) 2 (i+1) 5 (2 + (i +1) 5) = 2 i 4 [2(i 1) (i 2)] = 2 i 4 i, which completes the proof. Acknowledgements The authors would like to thank Noam Elkies, who suggested the particular way of counting the number of summands used in Theorem 6. References [1] K. Alladi and V.E. Hoggatt, Jr., Compositions with Ones and Twos, Fibonacci Quarterly 13 (1975), No. 3, [2] P. J. Cameron, Sequences Realized by Oligomorphic Permutation Groups, Journal of Integer Sequences 3 (2000), #P [3] P. Z. Chinn, G. Coyler, M. Flashman and E. Migliore, Cuisinaire Rods Go to College, Primus II (1992), No. 2, [4] R. P. Grimaldi, Compositions with Odd Summands, Congressus Numerantium 142 (2000), [5] M. Ghallab, A. Howe, et. al, PDDL - The Planning Domain Definition Language, Version 1.2. Technical Report CVC TR /DCS TR- 1165, Yale Center for Computational Vision and Control, 1998 [6] W. Lang, On Generalizations of the Stirling Number Triangles, Journal of Integer Sequences 3 (2000), # [7] Sloane s On-Line Encyclopedia of Integer Sequences, electronically published at njas/sequences 14